Vectors, n-tuples, and headaches oh my

  • Thread starter Thread starter kashiark
  • Start date Start date
  • Tags Tags
    Vectors
kashiark
Messages
210
Reaction score
0

Homework Statement


An n-tuple C has two representation in terms of the linearly independent pair A,B
C = (x-3)A - YB
C = -yA - (x+2)B

Solve for x and y.

Homework Equations


aA+bB+cC = 0
where A,B,C are linearly dependent n-tuples and a,b,c are real numbers that don't all equal 0

The Attempt at a Solution


I tried this problem about a billion different ways; however, none yielded promising results. I know the answer, and it is numerical! I don't want x and y in terms of A and B; they have numerical values. It would be helpful to explain what you're doing as you do it, but I can probably figure it out if you just show the work; laziness is something that I understand well. :smile:
 
Physics news on Phys.org
I'm not sure I get why you are confused. If A and B are linearly independent and (x-3)A-yB=-yA-(x+2)B, doesn't that mean (x-3)=(-y) and (-y)=(-(x+2))? That's two equations in two unknowns. Just solve them.
 
I guess I was over-thinking it; thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top